Question

If the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?​

Answer 1

Hey Buddy...

Here's your answer..

Step-by-step explanation:

Given

an = 0, n = ?

3rd term = 4

⇒ a + (3-1)d = 4 → 1

9th term = 73

⇒ a + (9-1)d = -8 → 2

By subtracting the both equations we will get ‘d’

(a +2d) – (a+8d) = 4 – (-8)

-6d = 12

d = -2

By substituting “d” in equation 1

a +2d = 4

a + (-2)2 = 4

a = 8

an = a+(n-1)d

0 = 8+(n-1)(-2)

-8 = (n-1)(-2)

4 = n-1

n = 5

∴ 0 is the 5th term in the series.

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Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

$\textbf{\underline{First\;term\;be\;p}}$

$\textbf{\underline{Common\; Difference\;be\;d}}$

Hence,

$\tt{\rightarrow p_{3}=4}$

p + 2d = 4 ..... (1)

$\tt{\rightarrow p_{9}=-8}$

p + 8d = -8 ..... (2)

${\boxed{\sf\:{Subtracting\;(1)\;from\;(2)}}}$

6d = -12

$\tt{\rightarrow d=\dfrac{-12}{6}}$

d = -2

${\boxed{\sf\:{Put\;the\;value\;of\;d\;in\;(1)}}}$

p + 2(-2) = 4

p + (-4) = 4

p - 4 = 4

p = 4 + 4

p = 8

$\textbf{\underline{First\;Term=8}}$

$\textbf{\underline{Common\; Difference=-2}}$

$\tt{\rightarrow p_{n}=0}$

p + (n - 1)d = 0

8 + (n - 1)(-2) = 0

n - 1 = 4

n = 4 + 1

n = 5

$\Large{\boxed{\sf\:{Hence\;5^{th}\;term\;is\;0}}}$